Mathematica 9 is now available
Services & Resources / Wolfram Forums
-----
 /
MathGroup Archive
2004
*January
*February
*March
*April
*May
*June
*July
*August
*September
*October
*November
*December
*Archive Index
*Ask about this page
*Print this page
*Give us feedback
*Sign up for the Wolfram Insider

MathGroup Archive 2004

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: Re: Proving inequalities with Mathematica

  • To: mathgroup at smc.vnet.net
  • Subject: [mg52553] Re: [mg52499] Re: [mg52491] Proving inequalities with Mathematica
  • From: DrBob <drbob at bigfoot.com>
  • Date: Wed, 1 Dec 2004 05:59:21 -0500 (EST)
  • References: <200411290622.BAA27997@smc.vnet.net> <200411301024.FAA01299@smc.vnet.net>
  • Reply-to: drbob at bigfoot.com
  • Sender: owner-wri-mathgroup at wolfram.com

I wish I'd thought of that. Brilliant, Andrzej.

FWIW, here's a process for making other examples (now that you've pointed the way):

Clear[a,b]
a[n_]=(100/99)^-n;
Simplify[Abs[a[n+1]/a[n]]<1,n>1]

True

(If that fails, choose another series.)

Now choose a[1] and a[2]:

s = Simplify[Sum[a[n], {n, 3, Infinity}]];
{a[1], a[2]} = {x, y} /. First@FindInstance[{x + y == -s, Abs@x > Abs@y >
   Abs@a@3}, {x, y}, Reals]

{-(490001/5000), 9703/10000}

All premises are guaranteed at this point (setting b == a), but the conclusion can NEVER be true, since

Sum[a[i]^2, {i, 1, Infinity}] < a[1]^2

is equivalent to:

Sum[a[i]^2, {i, 2, Infinity}] < 0

But wait -- could FindInstance fail? Consider this, just for s >= 0:

s =. ;
Reduce[{s >= 0, x + y == -s,
    Abs[x] > Abs[y] > Abs[a3],
    a3 \[Element] Reals}, {x, y}, Reals]
s > 0 && ((a3 <= -(s/2) &&
     x < a3 - s) ||
    (Inequality[-(s/2), Less, a3,
      LessEqual, 0] &&
     (x < a3 - s || -a3 - s < x <
       -(s/2))) ||
    (0 < a3 < s/2 &&
     (x < -a3 - s || a3 - s < x <
       -(s/2))) || (a3 >= s/2 &&
     x < -a3 - s)) && y == -s - x

The Or in the middle is a run-down of all the ways a3 can stand in relation to +-s/2 and 0; for every case we have a non-empty range in which to choose x and a formula for y. The situation is similar for s<0.

Bobby

On Tue, 30 Nov 2004 05:24:01 -0500 (EST), Andrzej Kozlowski <akoz at mimuw.edu.pl> wrote:

>
> On 29 Nov 2004, at 15:22, Toshiyuki (Toshi) Meshii wrote:
>
>> Hi,
>>
>> I was wondering whether Mathematica is useful for proving a problem of
>> inequality.
>> My problem is as follows:
>>
>> Let An and Bn (n=1,2,3...) be real sequences.
>> Some characteristics of these sequences are known.
>>
>> i) Abs[An+1/An] < 1
>> ii) Abs[Bn+1/Bn] < 1
>> iii) Sum[An, {1,Infinity}] = 0
>> iv) Sum[An*Bn, {1, Infinity}] = alpha (note: a real number)
>>
>> Then I want to prove with Mathematica that
>> 0 < Abs[alpha] < Abs[A1*B1]
>>
>> Does anyone have an idea?
>>
>> -Toshi
>>
>
> First of all, you have written your condition as
>
>> Abs[An+1/An] < 1
>
> Presumably you meant n+1 to be a subscript (if not the condition is
> always false).
> Assuming that,  Mathematica can certainly be helpful in proving your
> claim to be false. Here is an example.
>
> We define the sequence A[n] as follows:
>
> A[1] = -1; A[2] = Log[2];
> A[n_] = (-1)^(n - 1)/(n - 1);
>
> You can check that the condition Abs[A[n+1]/A[n]]<1 is always
> satisfied. We also define the sequence B[n] by
> B[i_] := A[i]
> so that the condition
>
> Abs[B[n+1]/B[n]]<1
>
> is satisifed.
>
> We check that (iii) is satisfied:
>
>
> A[1]+A[2]+Sum[A[n],{n,3,Infinity}]//Simplify
>
> 0
>
> Let's now compute alpha:
>
>
> alpha=A[1]*B[1]+A[2]*B[2]+Sum[A[i]*B[i],{i,3,Infinity}]//N
>
>
> 2.12539
>
> Let's see if your inequality holds:
>
>
> Abs[alpha]<Abs[A[1]*B[1]]
>
> False
>
> Oops ...
>
>
> Andrzej Kozlowski
> Chiba, Japan
> http://www.akikoz.net/~andrzej/
> http://www.mimuw.edu.pl/~akoz/
>
>
>
>



-- 
DrBob at bigfoot.com
www.eclecticdreams.net


  • Prev by Date: Re: New User - Programming
  • Next by Date: Re: New User - Programming
  • Previous by thread: 4th subharmoic functions
  • Next by thread: Boltzmann Exponent